Black and white.

Level pending

An urn contains black and white balls only. It is known that we can draw at random with equal probability two balls of same colour or two balls of different colours. What is the total number of balls in the urn if there are atleast 2014 balls and atmost 2114 balls.


The answer is 2025.

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1 solution

Suppose there are N N balls in the urn, b b of which are black, w w white. Then the given condition means that

( b 2 ) + ( w 2 ) = ( b 1 ) ( w 1 ) \dbinom{b}{2} + \dbinom{w}{2} = \dbinom{b}{1} \dbinom{w}{1}

b ( b 1 ) 2 + w ( w 1 ) 2 = b w \Longrightarrow \dfrac{b(b - 1)}{2} + \dfrac{w(w - 1)}{2} = bw

b 2 b + w 2 w = 2 b w \Longrightarrow b^{2} - b + w^{2} - w = 2bw

b 2 2 b w + w 2 = b + w = N ( b w ) 2 = N . \Longrightarrow b^{2} - 2bw + w^{2} = b + w = N \Longrightarrow (b - w)^{2} = N.

So N N must be a perfect square between 2014 2014 and 2114 2114 . The only possibility is N = 4 5 2 = 2025 N = 45^{2} = \boxed{2025} . This would make

b w = 45 , b + w = 2025 2 b = 2070 b = 1035 , w = 990 b - w = 45, b + w = 2025 \Longrightarrow 2b = 2070 \Longrightarrow b = 1035, w = 990

or b = 990 , w = 1035 b = 990, w = 1035 , if we had taken w b = 45 w - b = 45 .

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